/sci/ - Science & Math

[Anonymous No.3021173]

I'm having trouble with an algebra problem. This is it.


12 ( ( 5/6 + 3/4) - 1/6)
__________________
5 √ 2 (3)^2 + 7

My answer comes out to = 12(8+9) / 25

The book claims the answer is 17/25, though.

I have no idea what they did with the number 12. Can someone clarify this?

[Anonymous No.3021178]

>algebra

[Anonymous No.3021187]

>mfw OP can't do simple arithmatic
>mfw I can't spell arithmatic

[Anonymous No.3021188]

>>3021178

I only passed algebra 1 in high school and I feel guilty about it, so I got a book from the library to teach me algebra 2.

anyway can someone help me?

[Anonymous No.3021200]

how the fuck did you turn ( ( 5/6 + 3/4) - 1/6) into 8 + 9?

[Anonymous No.3021202]

That's not even algebra. And use LaTeX and I might consider helping you.

[Anonymous No.3021205]

Anyone want to help me out?

[Anonymous No.3021212]

<span class="math">\displaystyle \frac{120/12}{25}[/spoiler]

[Anonymous No.3021213]

>>3021205

>>3021200

[Anonymous No.3021218]

>>3021200

"Use the Commutative and associative properties to rearrange the fractions in the numerator;"

is what the book said

[Anonymous No.3021221]

>>3021212

This assumes:

__________________
5 √ 2 (3)^2 + 7

= <span class="math">5*\sqrt{2*3^2 + 7}[/spoiler]

[Anonymous No.3021235]

[Anonymous No.3021237]

>>3021218
well you better check yo shit cuz ( ( 5/6 + 3/4) - 1/6) = 34/24

[Anonymous No.3021241]

Can someone explain what they did with the 12 in the numerator? My answer would be right according to the book, if I could get rid of the 12, because my current answer comes out to 12(8+7) / 25

[Anonymous No.3021253]

Fucking write out your expression because it looks like shit at the moment.

<span class="math">\displaystyle \frac{12[(\frac{5}{6} + \frac{3}{4}) - \frac{1}{6}]}{5*\sqrt{2}*3^9 + 7}[/spoiler]?

<span class="math">\displaystyle \frac{12[(\frac{5}{6} + \frac{3}{4}) - \frac{1}{6}]}{5*\sqrt{2*3^9 + 7}}[/spoiler]

<span class="math">\displaystyle \frac{12[(\frac{5}{6} + \frac{3}{4}) - \frac{1}{6}]}{OP has a small penis}[/spoiler]?

[Anonymous No.3021265]

>>3021241
>Can someone explain what they did with the 12 in the numerator?
See:
>>3021235

What don't you get?

[Anonymous No.3021276]

>>3021265

What don't you get?

[Anonymous No.3021281]

>>3021276
I'm not the OP. Suffice it to say: I get it.

[Anonymous No.3021297]

>>3021281

Circle one.

[Anonymous No.3021304]

thanks a billion guys, aside from all the trolling. The guy that posted the pic showed me the way its done.

[Anonymous No.3021306]

>>3021297
No idea what you're talking about. The numerator is 17 and I've shown you how to arrive at it.

[Anonymous No.3021308]

>>3021304
Great. :) Cheers. And yeah, you have to filter out the idiots when you come here.

[Anonymous No.3021350]

>>3021306

Hurr I don't know maybe one of the equations six posts above yours:

>>3021253

It's probably safe to assume C.

[Anonymous No.3021578]

Having some difficulty with another one

Simplify y^-3 (y^2)^4 * y^4/yy^3

I came out with: y^8 * y^4/y^4

I assumed the y^4 fraction would cancel itself out, and the answer would be y^8 but I was wrong apparently.

the book states the answer is y^5

Can someone walk me through this one?

[Anonymous No.3021609]

>>3021578
Anyone wanna help?

[Anonymous No.3021632]

>>3021609

anyone feel like showing me how its done

[Anonymous No.3021661]

>>3021632
Sure, just give me a few minutes to make it look pretty...

[Anonymous No.3021665]

>>3021661

alright

[Anonymous No.3021694]

>>3021632
Simplify <span class="math"> \displaystyle{ y^{-3} (y^2)^4 \frac{ y^4}{y(y^3)} } [/spoiler]

<span class="math"> \displaystyle{ y^{-3} (y^2)^4 \frac{ y^4}{y(y^3)} = \frac{y^8}{y^3} (1) = y^5 } [/spoiler] 

[Anonymous No.3021706]

>>3021694
Let me know if I did too many steps at once; I'll be happy to show a bit more work/explain the steps if you don't understand. :)

[Anonymous No.3021720]

>>3021694

thanks

[Anonymous No.3021733]

>>3021706

That was good, I spotted my mistake.